Statics-Two wires holding traffic light

[lrp3395]

I dont know what i did wrong

Find the tension in the two wires supporting the traffic light
Mass of light= 33 kg
θ of FT1= 53°
θ of FT2=37°

Ʃfx=0
Ft1cos53-ft2cos37=0
ft1cos53=ft2cos37

Ʃfy=mg
ft1sin53-ft2sin37=mg
ft1sin53=mg+ft2sin37
ft1=(ft2sin37+mg)/sin53
ft1=1.33ft2

1.33ft2=(ft2sin37+mg)/sin53
1.33ft2=.72ft2+mg
.61ft2=323.73 N
Ft2=530.7n

ft1cos53=ft2cos37
ft1cos53=423.84n
ft1=704.26n

Homework Statement

Homework Equations

The Attempt at a Solution

 

[PhanthomJay]

Tension forces always pull away from the objects on which they act. In the horizontal direction, your equation is correct. Not so in the vertical direction.

 

[ssnapier]

This is what I got:
TLsin(53o) + TRsin(37o) - 294 N= 0 TRcos(37o) - TLcos(53o) = 0
TR = TLcos(53o)/cos(37o)

TLsin(53o) + TRsin(37o) - 294N= 0
TLsin(53o) + {TLcos(53o)/cos(37o)}sin(37o) - 294 = 0
TLsin(53o) + TLcos(53o)tan(37o) = 294N
TL(sin(53o) + cos(53o)tan(37o)) = 294N
TL = (294 N)/(sin(53o) + cos(53o)tan(37o)) = 234.79884N = 230N

TR = TLcos(53o)/cos(37o)
TR = (234.79884 N)cos(53o)/cos(37o) = 176.9336 N = 180N

Left wire has 230N of tension
Right wire has 180N of tension

 

[DeShark]

I don't know where you got the 294N from? 33kg * 9.8 = 323N. g is 9.8, not 8.9!

Your maths seems right though.

You can check your answer easily also. 230sin(53) + 180sin(37) = 292N, which is close enough (since you rounded your answers) to 294.

 

[ssnapier]

Well dangit... I guess I got that one wrong then!

 

[Bill Nye Tho]

PhantomJay is correct

Hope this helps.